What is the solution for y^{''}+y=x^2+1,y(0)=2,y(1)=0 ?

The solution for y^{''}+y=x^2+1,y(0)=2,y(1)=0 is y=3cos(x)-(3cos(1)sin(x))/(sin(1))+x^2-1

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y^{''}+y=x^2+1,y(0)=2,y(1)=0

y^{^{''}} + y = x^{2} + 1, y (0) = 2, y (1) = 0

y = 3 cos (x) - \frac{3 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

Solution steps

y^{''} (x) + y = x^{2} + 1

Solve linear ODE:

y = 3 cos (x) - \frac{3 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

y = 3 cos (x) - \frac{3 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 3 x - 4

\frac{d ^{2} x}{d t ^{2}} + 9 x = e^{t} + sin (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{2} e^{2 t} + e^{2 t}

y^{^{''}} - y = (1 - e^{2} x)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 2 sin (4 x)

What is the solution for y^{''}+y=x^2+1,y(0)=2,y(1)=0 ?
The solution for y^{''}+y=x^2+1,y(0)=2,y(1)=0 is y=3cos(x)-(3cos(1)sin(x))/(sin(1))+x^2-1